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Line 446:
={} &\begin{pmatrix}\alpha^{7}+ \alpha^{-3}x & 1\\ 1 & 0\end{pmatrix}
\begin{pmatrix}\alpha^4 + \alpha^{-5}x & 1\\ 1 & 0\end{pmatrix} \\
&\qquad \begin{pmatrix}\alpha^{-7}+ \alpha^{4}x+ \alpha^{-1}x^2+ \alpha^{6}x^3+ \alpha^{-1}x^4+ \alpha^{5}x^5\\ \alpha^{-3} +\left(\alpha^{-7}+ \alpha^{3}\right)x+ \left(\alpha^{3}+ \alpha^{-1}\right)x^2+ \left(\alpha^{-5}+ \alpha^{-6}\right)x^3+ \left(\alpha^3+ \alpha^{1}\right)x^4+ 2\alpha^{-6}x^5+ 2x^6\end{pmatrix} \\ [6pt]
={} &\begin{pmatrix}\left(1+ \alpha^{-4}\right)+ \left(\alpha^{1}+ \alpha^{2}\right)x+ \alpha^{7}x^2 & \alpha^{7}+ \alpha^{-3}x \\ \alpha^4+ \alpha^{-5}x & 1\end{pmatrix}
\begin{pmatrix}\alpha^{-7}+ \alpha^{4}x+ \alpha^{-1}x^2+ \alpha^{6}x^3+ \alpha^{-1}x^4+ \alpha^{5}x^5\\ \alpha^{-3}+ \alpha^{-2}x+ \alpha^{0}x^2+ \alpha^{-2}x^3+ \alpha^{-6}x^4\end{pmatrix} \\ [6pt]
={} &\begin{pmatrix}\alpha^{-3}+ \alpha^{5}x+ \alpha^{7}x^2 & \alpha^{7}+ \alpha^{-3}x \\ \alpha^4+ \alpha^{-5}x & 1\end{pmatrix}
\begin{pmatrix}\alpha^{-5}+ \alpha^{-4}x & 1\\ 1 & 0 \end{pmatrix} \\
&\qquad \begin{pmatrix}\alpha^{-3}+ \alpha^{-2}x+ \alpha^{0}x^2+ \alpha^{-2}x^3+ \alpha^{-6}x^4\\ \left(\alpha^{7}+ \alpha^{-7}\right)+ \left(2\alpha^{-7}+ \alpha^{4}\right)x+ \left(\alpha^{-5}+ \alpha^{-6}+ \alpha^{-1}\right)x^2+ \left(\alpha^{-7}+ \alpha^{-4}+ \alpha^{6}\right)x^3+ \left(\alpha^{4}+ \alpha^{-6}+ \alpha^{-1}\right)x^4+ 2\alpha^{5}x^5\end{pmatrix} \\ [6pt]
={} &\begin{pmatrix}\alpha^{7}x+ \alpha^{5}x^2+ \alpha^{3}x^3 & \alpha^{-3}+ \alpha^{5}x+ \alpha^{7}x^2\\ \alpha^{3}+ \alpha^{-5}x+ \alpha^{6}x^2 & \alpha^4+ \alpha^{-5}x\end{pmatrix}
\begin{pmatrix}\alpha^{-3}+ \alpha^{-2}x+ \alpha^{0}x^2+ \alpha^{-2}x^3+ \alpha^{-6}x^4\\ \alpha^{-4}+ \alpha^{4}x+ \alpha^{2}x^2+ \alpha^{-5}x^3\end{pmatrix}.
Line 458:
We have reached polynomial of degree at most 3, and as
:<math>\begin{pmatrix}-\left(\alpha^4+ \alpha^{-5}x\right) & \alpha^{-3}+ \alpha^{5}x+ \alpha^{7}x^2\\ \alpha^{3}+ \alpha^{-5}x+ \alpha^{6}x^2 & -\left(\alpha^{7}x+ \alpha^{5}x^2+ \alpha^{3}x^3\right)\end{pmatrix} \begin{pmatrix}
we get
:<math>
\begin{pmatrix}S(x)\Gamma(x)\\ x^6\end{pmatrix} = \begin{pmatrix}
Therefore,
Line 469:
:<math>S(x)\Gamma(x)\left(\alpha^{3} + \alpha^{-5}x + \alpha^{6}x^2\right) - \left(\alpha^{7}x + \alpha^{5}x^2 + \alpha^{3}x^3\right)x^6 = \alpha^{-4} + \alpha^{4}x + \alpha^{2}x^2 + \alpha^{-5}x^3.</math>
Let <math>\Lambda(x) = \alpha^{3}+ \alpha^{-5}x+ \alpha^{6}x^2.</math> Don't worry that <math>\lambda_0\neq 1.</math> Find by brute force a root of <math>\Lambda.</math> The roots are <math>\alpha^2,</math> and <math>\alpha^{10}</math> (after finding for example <math>\alpha^2</math> we can divide <math>\Lambda</math> by corresponding monom <math>\left(x - \alpha^2\right)</math> and the root of resulting monom could be found easily).
Let
Line 475:
:<math>\begin{align}
\Xi(x) &= \Gamma(x)\Lambda(x) = \alpha^3 + \alpha^4x^2 + \alpha^2x^3 + \alpha^{-5}x^4 \\
\Omega(x) &= S(x)\Xi(x) \equiv \alpha^{-4} + \alpha^4x + \alpha^2x^2 + \alpha^{-5}x^3 \bmod{x^6}
\end{align}</math>
Line 485:
:<math>\begin{align}
e_1 &=-\frac{\Omega(\alpha^4)}{\Xi'(\alpha^{4})} = \frac{\alpha^{-4}+\alpha^{-7}+\alpha^{-5}+\alpha^{7}}{\alpha^{-5}} =\frac{\alpha^{-5}}{\alpha^{-5}}=1 \\
e_2 &=-\frac{\Omega(\alpha^7)}{\Xi'(\alpha^{7})} = \frac{\alpha^{-4}+\alpha^{-4}+\alpha^{1}+\alpha^{1}}{\alpha^{1}}=0 \\
e_3 &=-\frac{\Omega(\alpha^{10})}{\Xi'(\alpha^{10})} = \frac{\alpha^{-4}+\alpha^{-1}+\alpha^{7}+\alpha^{-5}}{\alpha^{7}}=\frac{\alpha^{7}}{\alpha^{7}}=1 \\
e_4 &=-\frac{\Omega(\alpha^{2})}{\Xi'(\alpha^{2})} = \frac{\alpha^{-4}+\alpha^{6}+\alpha^{6}+\alpha^{1}}{\alpha^{6}}=\frac{\alpha^{6}}{\alpha^{6}}=1
\end{align}</math>
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